Please guide me in this problem. I am confused about whether its asking that having the relation $z>0$ does not satisfy the order axioms.
Any help would be really appreciated.
Thanks!
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2Hint: squares are nonnegative in an ordered field. – Math Gems Feb 23 '13 at 18:58
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so you mean that if z = i (then z>0) but in this case the square of z is -1 which is <0. Would that be enough to prove the above? – uh1 Feb 23 '13 at 19:10
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1Yes, assuming that you have proved said statement from the axioms that you are using. – Math Gems Feb 23 '13 at 19:15
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you mean if I have proved that -1<0? Yes, i have already established that. Thanks! – uh1 Feb 23 '13 at 19:25
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No, I mean that you need to prove that squares are nonnegative using whatever ordered field axioms that you are using (there are at least several variations). – Math Gems Feb 23 '13 at 19:29
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Well according to one of the axioms, $x.y\in P$ where $x,y\in P$ so if x = y then its quite obvious from this axiom that the squares are nonnegatives in ordered fields. What do you think? – uh1 Feb 23 '13 at 20:09
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Yes, that works fine. – Math Gems Feb 23 '13 at 20:11
2 Answers
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What they are asking is to show that no relation $<$ can exist that complies with the order axioms, i.e.:
- Only one of $a < b$, $a = b$, or $a > b$ is true
- If $a < b$ and $b < c$, then $a < c$
- If $a < b$ and $c < d$, then $a + c < b + d$
- If $0 < a < b$ and $0 < c < d$, then $a c < b d$
In this case, if we take $i > 0$ we get $i^2 = -1 < 0$, contradicting (4). So by (1) it must be $i < 0$. But $i^4 = 1 > 0$, again contradicting (4). So no relation $<$ can exist on $\mathbb{C}$ which complies with (1) to (4).
vonbrand
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Thanks for your answer. I have solved it by evoking the contradiction by letting $z=i$ and then stating that the square of z is $z.z =-1<0$ whereas squares must be nonnegative in ordered rings according to one of the order axioms (which says x.y∈P where x,y∈P so if x = y then its quite obvious from this axiom that the squares are nonnegatives in ordered fields) – uh1 Feb 23 '13 at 20:48
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@uh1, the same as mine with a shortcut. Great! But please close the issue accepting an answer (or providing your own, and accepting that). – vonbrand Feb 23 '13 at 20:50
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3it is not true that: if a < b and c < d then ac < bd a = -10, b = -8, c = -6, d < -4: (-10)(-6) < (-8)(-4), 60 < 32 CLEARLY FALSE – May 01 '13 at 15:55
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1user75388 The < operator should change sign to > IF multiplied by negative numbers (also in this case). – Theodor Johnson Oct 03 '17 at 13:50
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The fourth axiom should be $\forall a,b,c \in \mathbb R$ ,if $ a<b$ and $0<c$ then $ac<bc$. – john Sep 16 '20 at 07:59
